Question

A particle is projected upwards from level ground at an angle with horizontal such that its range is twice the maximum height reached by it.The angle made by its velocity vector with the horizontal at ahorizontal distance R/4 from the point of projection(R is horizontal range)?

Answer 1

45 degrees?

Answer 2

Your answer is in this picture

Answer 3

s @anusha.p u r corrct but how?

Answer 4

d=(v^2/g)*sin2theta

Answer 5

@an can u solve it

Answer 6

@shahzadjalbani can u help??

Answer 7

In what

Answer 8

by telling how to do??

Answer 9

@lalaly

Answer 10

@Mertsj

Answer 11

Hey @shakir I'm not really great at this stuff and I'd probably make you fail but @parthkohli is really good with math stuff :)

Answer 12

Or! @blues she's not on right now but I heard she's a genius when it comes to this type of stuff :3

Answer 13

Velocity is not given ?

why there are two unknown variables in your question :P ?

Answer 14

it is give R=2Hmax

Answer 15

and what about R/4

Answer 16

so R/4 = Hmax/2

Answer 17

Good,
so did you try compare them ?

Answer 18

s but not getting

Answer 19

me too ;)

Answer 20

Firstly let's express what we need to find in formulas

Answer 21

v0y = v0 sin alpha
v0x = v0 cos alpha
so
v0y/v0x = tan aplha
so if you will get values of v0y and v0x at the R/4 point
you will easily find the aplha

Answer 22

v0y is vertical component of initial velocity
v0x is horizontal component of initial velocity

Answer 23

vox is constant so its value doesn't change through all the trajectory and if Beta is initial angle made by velocity vector the its value is
vox = vo cos Beta

Answer 24

ok

Answer 25

Beta is the angle made by Vo and horizon

Answer 26

So you need to find only voy component at R/4 point

Answer 27

@ArchiePhysics both angle and initial velocity are not give so how???

Answer 28

@anusha.p please don't confuse aplha and the other initial angle
Aplha is changing angle so its variable
And let Beta be the second angle i.e. initial angle so it is some constant value.

Answer 29

@shakir you don't have a choice you have to use initial angle in your formula or let's see if we can cancel it out

Answer 30

at R/4 the time interval value is t = t(total)/4 = R/(4vox)

Answer 31

Ok @anusha.p it is just the matter of convention I said that just to warn that you have to distinct between two angles. One of them is variable the second is constant.

Answer 32

i gave up!!!!!!!

Answer 33

There are some equations that you can use for solving this problem. I will make new designations:
T = t(total)/2 - time for a particle to rise the top of trajectory or to fall from it.
R - range, I think other letters is familiar to you, here

R = vox 2 T
Hmax = voy T - gT^2/2
voy = gT
using previous two equations we get
Hmax = gT^2/2
Let's solve that together

Answer 34

R = vox 2 T?? why

Answer 35

R= u cos theta *T

Answer 36

R = vox t(total) <= do you agree with this?
T = t(total)/2

Answer 37

yes

Answer 38

Ok I found that
v0y/vox = 2
alpha = arctan 2
but it is not correct. Let's see together what I got, may be you will find mistakes

Answer 39

u-velocity vth wich the object is thrown.
ux=initial horizontal velocity=u cos q,where q is the angle of projection.
uy=u sin q=initial vertical velocity
vx=ux(at any time t since aceleration in horizontal direction is zero)
vy=uy-g*t
tan t=vy/vx ,t is the angle made by the velocity vector horizontally.

Answer 40

from previous equations I derived following
v0x = R/2T
R = 2 Hmax
Hmax = gT^2/2
2Hmax = gT^2
R = gT^2
vox = gT^2/2T = gT/2
voy = gT
voy/vox=gT/gT/2=2

Answer 41

I think I got it

Answer 42

the time is t = T/2

Answer 43

I got theta =45 degree, but without using R/4
is it wrong ?

Answer 44

now,given range=2*max height
( u^2sin2q)/g =2*(usinq)^2 /2g.
u get tanq=2.(angle of projection found)

now,calculating the time required to reach R/4
ucosq=(R/4)/t.(since h.velocity is const)
t=R/(4ucosq)
gt=usinq/2(by replacing the range formula)
Vy=uy-gt=usinq-usinq/2=usinq/2
tant=vy/vx=usinq/(2*ucosq)=tanq/2=1
therefore t=45 degrees!!!!!!!!!

Answer 45

Vy(T/2) = Voy - g(T/2) = gT/2 - vertical component of velocity at R/4 or at the moment of time T/2 use that instead, so now we get 1 for tan alpha
Vy(T/2)/Vox = 1

Answer 46

The right answer is 45 degrees as
arctan 1 = 45 degrees